Solution
= Solution
The real <logarithm> restricts the approach to $x\to0^+$. The <Taylor expansion> $\sin x-x=-x^3/6+O(x^5)$ gives
$$
\frac1x+\frac{\ln x}{\sin x-x}
=\frac{-6\ln x}{x^3}\bigl(1+O(x^2)\bigr)+\frac1x.
$$
The positive first term dominates, since $x^2/|\ln x|\to0$. Thus the requested extended <limit> is $\boxed{+\infty}$. There is no two-sided real <limit> across zero because the real <logarithm> is undefined for $x<0$.